令F0=1,F1=1,Fk=Fk-1+Fk-2,即Fk为斐波那契数列.试证明:Fi≤FjF(i-j)+F(j+1)F(i-j-1),这里i≥j+1∈Z+
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![令F0=1,F1=1,Fk=Fk-1+Fk-2,即Fk为斐波那契数列.试证明:Fi≤FjF(i-j)+F(j+1)F(i-j-1),这里i≥j+1∈Z+](/uploads/image/z/10769162-50-2.jpg?t=%E4%BB%A4F0%3D1%2CF1%3D1%2CFk%3DFk-1%2BFk-2%2C%E5%8D%B3Fk%E4%B8%BA%E6%96%90%E6%B3%A2%E9%82%A3%E5%A5%91%E6%95%B0%E5%88%97.%E8%AF%95%E8%AF%81%E6%98%8E%EF%BC%9AFi%E2%89%A4FjF%28i-j%29%2BF%28j%2B1%29F%28i-j-1%29%2C%E8%BF%99%E9%87%8Ci%E2%89%A5j%2B1%E2%88%88Z%2B)
令F0=1,F1=1,Fk=Fk-1+Fk-2,即Fk为斐波那契数列.试证明:Fi≤FjF(i-j)+F(j+1)F(i-j-1),这里i≥j+1∈Z+
令F0=1,F1=1,Fk=Fk-1+Fk-2,即Fk为斐波那契数列.试证明:Fi≤FjF(i-j)+F(j+1)F(i-j-1),这里i≥j+1∈Z+
令F0=1,F1=1,Fk=Fk-1+Fk-2,即Fk为斐波那契数列.试证明:Fi≤FjF(i-j)+F(j+1)F(i-j-1),这里i≥j+1∈Z+
用数学归纳法.
证明j具有性质:对任意正整数i ≥ j+1都有Fi ≤ Fj·F(i-j)+F(j+1)·F(i-j-1).
若j = 0,Fi ≤ F0·Fi+F1·F(i-1) = Fi+F(i-1)显然对任意i ≥ j+1 = 1成立.
若j = 1,Fi ≤ F1·F(i-1)+F2·F(i-2) = F(i-1)+2F(i-2) = Fi+F(i-2)也对任意i ≥ j+1 = 2成立.
假设对j < k,Fi ≤ Fj·F(i-j)+F(j+1)·F(i-j-1)对任意i ≥ j+1成立.
则j = k时,对任意i ≥ j+1 = k+1,有i-1 ≥ k,i-2 ≥ k-1.由j = k-1,k-2时的归纳假设,有:
F(i-1) ≤ F(k-1)·F(i-k)+Fk·F(i-k-1),F(i-2) ≤ F(k-2)·F(i-k)+F(k-1)·F(i-k-1).
相加得Fi = F(i-1)+F(i-2) ≤ (F(k-1)+F(k-2))·F(i-k)+(Fk+F(k-1))·F(i-k-1) = Fk·F(i-k)+F(k+1)·F(i-k-1).
即j = k时,Fi ≤ Fj·F(i-j)+F(j+1)·F(i-j-1)也对任意正整数i ≥ j+1成立.
于是命题对任意自然数j成立,即对任意i ≥ j+1,有Fi ≤ Fj·F(i-j)+F(j+1)·F(i-j-1).